An electric field set up by an unknown charge distribution. U0 is the amount of

Question

An electric field set up by an unknown
charge distribution.
U0 is the amount of work needed to bring a
point charge of charge q0 in from infinity to a
point P .
If the charge q0 is returned to infinity, how
much work W would it take to bring a new
charge q = 4 q0 from infinity to point P ?


part 2
What is the electric potential V at point P ?


part 3
If point P in Part 1 is at a distance a along
the x-axis, how much work W would it take to
bring the charge q0 to a point
P? at a distance 4 a along the x-axis?

Explanation / Answer

You could assemble the charges one at a time but, since the E field forces add vectorially, the total work is the sum of works required to assemble all of the pairs. This is simplified by observing that all the charges have the same value q and the distances are integer multiples of the shortest distance r. Given q=60 uC and r = 1.4 cm, the work required to bring a pair of charges to a distance r apart is W1 = (q^2)*(4*pi*e0*r). The work required to bring a pair to a distance 2r apart is W2 = (W1)/2. (why?) And for a distance 3r, W3 = (W1)/3. In the final assembly, there are 3 pairs r apart, 2 pairs 2r apart and a single pair 3r apart. The total work is: W = 3*W1 + 2*W2 + W1 = (13/3)*W1

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